\beginsection 29.11

Show that $\sin x\le x$ for all $x\ge0$.
{\it Hint:} Show that $f(x)=x-\sin x$ is increasing on $[0,\infty).$

\medskip
$$f^\prime(x)=1+\cos x$$
Since the range of $\cos x$ is $[-1,1]$, the range of $f^\prime(x)$ is
$[0,2]$. By corollary 29.7 we conclude that $f$ is an increasing function
because $f^\prime(x)\ge0$ for all $x\in[0,\infty)$.
Since $f(0)=0$ and $f$ is an increasing function we must have
$f(x)\ge0$ for $x\in[0,\infty)$.
Therefore $x\ge\sin x$.